Can I state the Mandelbrot Set/Equation like this?

[emergentecon]

f(z) = z² + c

 

[Raghav Gupta]

But it must be under iteration. Then definitely you can do that.
Here I suppose you are saying z and c are complex numbers.

 

[emergentecon]

But it must be under iteration. Then definitely you can do that.
Here I suppose you are saying z and c are complex numbers.

Correct yes . . . under iteration, and z + c are complex numbers.
Thanks!

 

[HallsofIvy]

Just the simple equation "$f(z)= z^2+ c$" is not a "definition" of anything! The Mandelbrot set is the set of all points, c, in the complex plane such that the recursion "$z_{n+1}= z_n^2+ c$, $z_0= 0$"

 

[emergentecon]

Just the simple equation "$f(z)= z^2+ c$" is not a "definition" of anything! The Mandelbrot set is the set of all points, c, in the complex plane such that the recursion "$z_{n+1}= z_n^2+ c$, $z_0= 0$"

I can't say this helps me much.

Is it wrong to say that if you iterate the function f(z) = z^2 + c over the complex numbers, that you get the mandelbrot set?

 

[emergentecon]

Just the simple equation "$f(z)= z^2+ c$" is not a "definition" of anything! The Mandelbrot set is the set of all points, c, in the complex plane such that the recursion "$z_{n+1}= z_n^2+ c$, $z_0= 0$"

For instance, if I wanted to implement the recursion in Excel, I would in essence specify it as f(z) = z^2 + c

Just the simple equation "$f(z)= z^2+ c$" is not a "definition" of anything! The Mandelbrot set is the set of all points, c, in the complex plane such that the recursion "$z_{n+1}= z_n^2+ c$, $z_0= 0$"

The mandelbrot set is often 'summarised' as z = z^2 + c so why is it wrong to write f(z) = z^2 + c ????

 

[HallsofIvy]

I can't say this helps me much.

Is it wrong to say that if you iterate the function f(z) = z^2 + c over the complex numbers, that you get the mandelbrot set?

??? Just "iterating a function" will give you a lot of numbers and a lot of points where the iteration does not converge to any number. Again, the "Mandlebrot set" is the set of complex numbers for which the iteration does converge.

The mandelbrot set is often 'summarised' as z = z^2 + c so why is it wrong to write f(z) = z^2 + c ????

I don't know what you mean by "summarized" here. What do you do with that iteration to get the Mandlebrot set. (The "Julia sets" use that same iteration but in a different way.)

 

[stedwards]

Given f(z) = z² + c then z_n+1 = f(z_n). z₀=0, c ranges.

The mandelbrot set is over the domain, c.

If you want to produce a mandelbrot set, you need more than an equations. You need an algorithm.

 

[micromass]

The mandelbrot set is often 'summarised' as z = z^2 + c so why is it wrong to write f(z) = z^2 + c ????

It all depends on context. If you are talking to somebody about dynamical systems, then yes, you can write it like that. If you're not in the correct context, then no, you can't. It all depends on whether the other person will understand you or not.

 

[emergentecon]

It all depends on context. If you are talking to somebody about dynamical systems, then yes, you can write it like that. If you're not in the correct context, then no, you can't. It all depends on whether the other person will understand you or not.

I mis-stated my question, meant to say his equation, as opposed to the set.
As I know, when he was asked about his work, he wrote down the equation as: z -> z^2 + c
So was wondering if, in this context, z = z^2 + c or f(z) = z^2 + c as opposed to z(n+1) = z(n)^2 + c

 

[Mark44]

I mis-stated my question, meant to say his equation, as opposed to the set.
As I know, when he was asked about his work, he wrote down the equation as: z -> z^2 + c

I would bet that it was z <- z² + c, with the idea being that, starting with a specified complex number c and some complex number z, you square z, add c, and use that as your new z. Then repeat. And repeat. Ad infinitum.

So was wondering if, in this context, z = z^2 + c or f(z) = z^2 + c as opposed to z(n+1) = z(n)^2 + c

The latter formula better captures the idea of an endless sequence of complex numbers.

 

[emergentecon]

Thanks!